Exercise about wedge product and multilinear forms

Question

I'm considering $\omega\in \Lambda^{2q+1}(V^\ast)$, i.e. a multilinear skew-symmetric form.

I want to prove that $\omega\wedge\omega=0$.

How shall I proceed? Any suggestions?

Do I have to write $\omega$ as linear combination of $\epsilon_{i_1}\wedge\dots\epsilon_{i_q}$ first of all?

Do I loose generality if I suppose $\omega =\alpha \epsilon_{i_1}\wedge\dots\epsilon_{i_q}$?

Answer 1

There isn't much you need to do. The wedge product satisfies the relationship: $\alpha \wedge \beta = (-1)^{pq} \beta \wedge \alpha$ if $\alpha, \beta \in \Lambda^p, \Lambda^q$ respectively.

In your case $$\omega \wedge \omega = (-1)^{(2q+1)(2q+1)} \omega \wedge \omega = -\omega \wedge \omega$$ That can only happen if $\omega \wedge \omega = 0$.